Proof of Sums, Differences and Scalar Multiples of Functions Twice Differentiable at a Point

Throughout we use the notation of the statement. We record two identities used repeatedly: for $h\in\mathbb{R}^{n}$, claims 2, 4 and 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite give

$$(p+p')\cdot h=p\cdot h+p'\cdot h,\qquad (\mu p)\cdot h=\mu\,(p\cdot h),$$

and claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum together with claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite gives

$$h\cdot\bigl((B+B')h\bigr)=h\cdot(Bh)+h\cdot(B'h),\qquad h\cdot\bigl((\mu B)h\bigr)=\mu\,\bigl(h\cdot(Bh)\bigr).$$

Claim 1. Let $\varepsilon\in\mathbb{R}$ be positive. Then $\tfrac{\varepsilon}{2}$ is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field. By Twice Differentiability at a Point (i) applied to $f$ and to $g$ with the parameter $\tfrac{\varepsilon}{2}$, there are positive $\delta_{1},\delta_{2}\in\mathbb{R}$ such that every $h\in\mathbb{R}^{n}$ with $\lVert h\rVert<\delta_{1}$ satisfies $y+h\in U$ and

$$\Bigl|f(y+h)-f(y)-p\cdot h-\tfrac{1}{2}h\cdot(Bh)\Bigr|\le\tfrac{\varepsilon}{2}\lVert h\rVert^{2},$$

and every $h$ with $\lVert h\rVert<\delta_{2}$ satisfies $y+h\in U$ and the corresponding estimate for $g$, $p'$ and $B'$. Let $\delta$ be the smaller of $\delta_{1}$ and $\delta_{2}$, which exists by claim 9 of Elementary Order Arithmetic in an Ordered Field and is positive. Let $h\in\mathbb{R}^{n}$ with $\lVert h\rVert<\delta$; then $y+h\in U$ and both estimates hold. By the identities recorded above,

$$(f+g)(y+h)-(f+g)(y)-(p+p')\cdot h-\tfrac{1}{2}h\cdot\bigl((B+B')h\bigr)$$

is the sum of the two quantities whose absolute values are estimated, so by claim 5 of Properties of the Absolute Value in an Ordered Field its absolute value is at most $\tfrac{\varepsilon}{2}\lVert h\rVert^{2}+\tfrac{\varepsilon}{2}\lVert h\rVert^{2}=\varepsilon\lVert h\rVert^{2}$, the last equality by claim 8 of Elementary Order Arithmetic in an Ordered Field. As $\varepsilon$ was an arbitrary positive real number, $f+g$ is twice differentiable at $y$ with first-order coefficient $p+p'$ and Hessian $B+B'$.

Claim 2. Suppose first $\mu=0$. Then $\mu f$ is the function with constant value $0$, $\mu p$ is the origin of $\mathbb{R}^{n}$ and $\mu B=0_{n}$, the real $n\times n$ matrix all of whose entries are $0$, which lies in $\mathcal{S}(n)$. For every $h$ we have $(\mu p)\cdot h=0$ and $h\cdot\bigl((\mu B)h\bigr)=0$ by the identities recorded above, so the quantity to be estimated is $0$, and $0\le\varepsilon\lVert h\rVert^{2}$ for every positive $\varepsilon$; any positive $\delta$ with the property that $\lVert h\rVert<\delta$ implies $y+h\in U$ will do, and such a $\delta$ exists because $U$ is open and $y\in U$.

Suppose now $\mu\ne0$, so that $0<|\mu|$ by claim 1 of Properties of the Absolute Value in an Ordered Field and $|\mu|^{-1}$ exists and is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field. Let $\varepsilon$ be positive and apply Twice Differentiability at a Point (i) to $f$ with the positive parameter $\varepsilon\,|\mu|^{-1}$, obtaining a positive $\delta$ such that $\lVert h\rVert<\delta$ implies $y+h\in U$ and

$$\Bigl|f(y+h)-f(y)-p\cdot h-\tfrac{1}{2}h\cdot(Bh)\Bigr|\le\varepsilon\,|\mu|^{-1}\lVert h\rVert^{2}.$$

By the identities recorded above, the corresponding quantity for $\mu f$, $\mu p$ and $\mu B$ equals $\mu$ times the quantity for $f$, $p$ and $B$, so by claim 4 of Properties of the Absolute Value in an Ordered Field its absolute value is at most $|\mu|\,\varepsilon\,|\mu|^{-1}\lVert h\rVert^{2}=\varepsilon\lVert h\rVert^{2}$.

Claim 3. By claim 2 with $\mu=-1$, the function $(-1)g$ is twice differentiable at $y$ with first-order coefficient $(-1)p'$ and Hessian $(-1)B'$. The function $f-g$ is $f+(-1)g$, and $p+(-1)p'=p-p'$ and $B+(-1)B'=B-B'$, these being the definitions of the difference of points of $\mathbb{R}^{n}$ and of the difference of real matrices read entrywise. Claim 1 now gives the assertion.